Paper I — Q8
For an FM communication system with β = 2 and white channel noise with PSD Sₙ(ω) = 10⁻¹⁰, the output SNR is found to be 28 dB…
For an FM communication system with β = 2 and white channel noise with PSD Sₙ(ω) = 10⁻¹⁰, the output SNR is found to be 28 dB. The base band signal m(t) is Gaussian, band-limited to 15 kHz, and 3σ loading is used.
Determine the following :
The received signal power 'Sᵢ'
The output signal power 'S₀'
The output noise power 'N₀' 20 marks
A three-phase, 4-pole, 50 Hz induction motor has a rotor resistance of 4·5 Ω/phase and a standstill reactance of 8·5 Ω/phase with no external resistance in the rotor circuit. The starting torque of the motor is 85 Nm. Neglecting stator voltage drop, determine the following :
The rotor voltage at standstill
The starting torque, if a 3 Ω resistance were added in each rotor phase
The rotor induced voltage and the torque at a slip of 0·03 20 marks
A 220 V, 1500 rpm, 10 A separately excited dc motor has an armature resistance of 1 ohm. It is fed from a single phase fully-controlled bridge rectifier with an ac source voltage of 230 V, 50 Hz. Assuming continuous load current, determine the following :
Motor speed at the firing angle of 30° and torque of 5 Nm
Developed torque at the firing angle of 45° and speed of 1000 rpm 10 marks
हिंदी में प्रश्न पढ़ें
β = 2 तथा श्वेत चैनल रव जिसमें PSD Sₙ(ω) = 10⁻¹⁰ है, से संयुक्त एक FM संचार प्रणाली के लिए निर्गत SNR = 28 dB पाया गया है । आधार पटिका (बेस बैंड) संकेत m(t) गॉसियन है जो 15 kHz पर पटिका-सीमित है, तथा 3σ भारण (लोडिंग) का प्रयोग किया गया है ।
निम्नलिखित का मान ज्ञात कीजिए :
प्राप्त संकेत शक्ति 'Sᵢ'
निर्गत संकेत शक्ति 'S₀'
निर्गत रव शक्ति 'N₀' (20 अंक)
एक त्रि-कला, 4-ध्रुव, 50 Hz प्रेरण मोटर का घूर्णक प्रतिरोध 4·5 Ω/कला तथा गतिहीन अवस्था में, घूर्णक परिपथ में किसी बाह्य प्रतिरोध के बिना प्रतिघात 8·5 Ω/कला है । मोटर का प्रवर्तन बल-आघूर्ण 85 Nm है । स्टेटर के वोल्टता पात (गिरावट) को नगण्य मानते हुए, निम्नलिखित का मान ज्ञात कीजिए :
गतिहीन अवस्था में घूर्णक की वोल्टता
यदि घूर्णक की प्रत्येक कला में 3 Ω प्रतिरोध जोड़ा गया हो, तो प्रवर्तन बल-आघूर्ण
0·03 सर्पण (स्लिप) की अवस्था में घूर्णक की प्रेरित वोल्टता तथा बल-आघूर्ण (20 अंक)
एक 220 V, 1500 rpm, 10 A, अलग (पृथक) से उत्तेजित dc मोटर का आर्मेचर प्रतिरोध 1 ओम है । इसे एक 230 V के ac स्रोत वोल्टता, 50 Hz से संयोजित एक कला पूर्ण-नियंत्रित सेतु दिष्टकारी से भारित किया जाता है । भार धारा को सतत मानते हुए निम्नलिखित का मान ज्ञात कीजिए :
30° प्रसर्जन कोण तथा 5 Nm बल-आघूर्ण पर मोटर की गति
45° प्रसर्जन कोण तथा 1000 rpm गति पर विकसित बल-आघूर्ण (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(b) A logic circuit diagram labeled 'Figure 4(b)' at the bottom. The circuit consists of four 2-to-1 multiplexers on the left and one 4-to-1 multiplexer on the right. The four 2-to-1 multiplexers are arranged vertically. The top 2-to-1 MUX has inputs D0 (connected to input 0) and D1 (connected to input 1). The second 2-to-1 MUX has inputs D2 (connected to input 0) and D3 (connected to input 1). The third 2-to-1 MUX has inputs D4 (connected to input 0) and D5 (connected to input 1). The bottom 2-to-1 MUX has inputs D6 (connected to input 0) and D7 (connected to input 1). The select line for all four 2-to-1 multiplexers is connected to a common input labeled S0. The output of the top 2-to-1 MUX connects to input 00 of the 4-to-1 MUX. The output of the second 2-to-1 MUX connects to input 01 of the 4-to-1 MUX. The output of the third 2-to-1 MUX connects to input 10 of the 4-to-1 MUX. The output of the bottom 2-to-1 MUX connects to input 11 of the 4-to-1 MUX. The 4-to-1 MUX has two select lines labeled S2 and S1. The output of the 4-to-1 MUX is labeled Y.
(c) Circuit diagram of a three-phase bridge inverter (labeled Figure 6(b)). A DC voltage source Vs is connected to the inverter bridge. The bridge consists of six switching devices (transistors Q1 to Q6) arranged in three legs. Each leg has an upper switch (Q1, Q3, Q5) and a lower switch (Q4, Q6, Q2). Each switch has an anti-parallel diode (D1 to D6). The midpoints of the legs are labeled 'a', 'b', and 'c'. These three points connect to a star-connected (Y-connected) resistive load. Each phase of the load has a resistance labeled 'R'. The neutral point of the load is labeled 'n'. The text below the figure states: 'A three-phase bridge inverter shown in Figure 6(b) is used to feed a Y-connected resistive load with R = 10 ohm per phase. The dc input to the inverter Vs = 400 V and the output frequency is 50 Hz. If the inverter is operating with 180 degree conduction mode,' followed by four sub-questions: (i) compute the rms value of the load current, (ii) compute the rms value of the current in each switching device, (iii) calculate the output power, and (iv) draw the waveforms of phase and line voltages.
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) FM system powers Noise: the two-sided PSD is 𝒩/2 = 10⁻¹⁰, so 𝒩 = 2 × 10⁻¹⁰ W/Hz. Output SNR = 28 dB = 631. Above threshold, S₀/N₀ = 3β²(m²‾/m_p²)γ with γ = S_i/(𝒩B). With 3σ loading m_p = 3σ, so m²‾/m_p² = 1/9 and S₀/N₀ = 3 × 4 × (1/9)γ = (4/3)γ. (i) γ = 631 × 3/4 = 473.2, so S_i = γ𝒩B = 473.2 × 2 × 10⁻¹⁰ × 15 × 10³ = 1.42 × 10⁻³ W (1.42 mW). (ii) Take the discriminator output as the frequency deviation k_f m(t) (unit discriminator constant, since none is given). Then k_f m_p = Δω = 2πβB, so k_f σ = 2π × 2 × 15 × 10³/3 = 6.28 × 10⁴ rad/s and S₀ = (k_fσ)² = 3.95 × 10⁹ (rad/s)² (10⁸ Hz² in hertz units). (iii) The output noise PSD rises as ω²𝒩/A² with A² = 2S_i, so N₀ = ∫₋B^B (2πf)²(𝒩/A²) df = 4π²𝒩B³/(3S_i) = 4π² × 2 × 10⁻¹⁰ × (15 × 10³)³/(3 × 1.42 × 10⁻³) N₀ ≈ 6.26 × 10⁶ (rad/s)². Check: S₀/N₀ = 3.95 × 10⁹/6.26 × 10⁶ = 631 = 28 dB. Only S_i is an absolute power; S₀ and N₀ scale with the discriminator constant, but their ratio does not.
(b) Three-phase induction motor N_s = 120 × 50/4 = 1500 rpm, ω_s = 157.08 rad/s. With the stator drop neglected, the standstill rotor emf E₂ is constant and T = (3/ω_s) sE₂²R₂/(R₂² + (sX₂)²). (i) At s = 1: 85 = (3/157.08) E₂² × 4.5/(4.5² + 8.5²), so E₂² = 85 × 157.08 × 92.5/13.5 = 91 480 and E₂ ≈ 302.5 V per phase (about 524 V line-to-line if star-connected). (ii) With 3 Ω added, R₂′ = 7.5 Ω: T_st = (3/157.08) × 302.5² × 7.5/(7.5² + 8.5²) = 1747 × 7.5/128.5 ≈ 102 N·m (a 20% rise, since R₂ moves towards X₂). (iii) At s = 0.03 (no external resistance): the rotor induced emf is sE₂ = 0.03 × 302.5 = 9.07 V per phase at 1.5 Hz. Z₂ = 4.5 + j0.255 Ω = 4.507 Ω, so I₂ = 2.01 A and T = 3I₂²R₂/(sω_s) = 3 × 2.013² × 4.5/(0.03 × 157.08) ≈ 11.6 N·m.
(c) Dc motor on a single-phase fully-controlled bridge Rated E_b = 220 − 10 × 1 = 210 V at ω = 157.08 rad/s, so K = 210/157.08 = 1.337 V·s/rad and T = KI_a. Converter output V₀ = (2V_m/π)cos α = (2 × 325.3/π)cos α = 207.1 cos α. (i) α = 30°, T = 5 N·m: V₀ = 179.3 V; I_a = 5/1.337 = 3.74 A; E_b = 179.3 − 3.74 = 175.6 V; ω = 131.3 rad/s, so N ≈ 1254 rpm. (ii) α = 45°, N = 1000 rpm: V₀ = 146.4 V; E_b = 210 × 1000/1500 = 140 V; I_a = 146.4 − 140 = 6.4 A, so T = KI_a ≈ 8.6 N·m.
How this answer will be evaluated
What earns full marks: All three parts solved with correct formulas, clear steps, and proper units; equivalent circuits or models shown where appropriate.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Received signal power, output signal power and output noise power of an FM system with β = 2 and 3σ loading at 28 dB output SNR. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Convert 28 dB to 631 and use S₀/N₀ = 3β²(m²‾/m_p²)γ = (4/3)γ with 3σ loading, with 𝒩 = 2 × 10⁻¹⁰ W/Hz.
- Find S_i = γ𝒩B ≈ 1.42 mW.
- Find S₀ and N₀ (about 3.95 × 10⁹ and 6.26 × 10⁶ (rad/s)² for a unit discriminator) with S₀/N₀ = 631.
Loses marks
- Confusing S_n(ω) = 𝒩/2 with 𝒩 (missing factor of 2).
- Using the dB value directly as a ratio.
Earns more
- Explicit dB-to-ratio conversion.
- Correct two-sided PSD factor 𝒩/2 and the ω² noise spectrum at the discriminator.
Extra mark
- Transmission bandwidth 2(β + 1)B = 90 kHz mentioned.
- (b) Rotor standstill voltage, starting torque with 3 Ω added per phase and rotor emf and torque at 0.03 slip. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use T = (3/ω_s) sE₂²R₂/(R₂² + (sX₂)²) with ω_s = 157.08 rad/s and find E₂ ≈ 302.5 V per phase from T_st = 85 N·m.
- Find T_st = 102 N·m with R₂′ = 7.5 Ω.
- Find the induced emf 9.07 V and the torque 11.6 N·m at s = 0.03.
Loses marks
- Forgetting to square terms in the torque formula.
- Using the stator voltage instead of the rotor induced emf.
Earns more
- Assumption of constant rotor emf (stator drop neglected) stated.
- Rotor impedance R₂ + jsX₂ used at s = 0.03.
Extra mark
- Standstill rotor current 31.5 A.
- (c) Speed at α = 30° and 5 N·m, and torque at α = 45° and 1000 rpm for a separately excited dc motor on a single-phase fully-controlled bridge. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Find the motor constant K = 210/157.08 = 1.337 V·s/rad and V₀ = (2V_m/π)cos α.
- Apply V₀ = E_b + I_aR_a to get N ≈ 1254 rpm at α = 30°, T = 5 N·m.
- Get T ≈ 8.6 N·m at α = 45°, 1000 rpm (I_a = 6.4 A).
Loses marks
- Using the peak voltage instead of the average output voltage.
- Ignoring the armature resistance drop.
Earns more
- Continuous conduction assumed.
- E_b = Kω with the constant taken from rated values.
Extra mark
- Bridge circuit with the motor drawn.
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
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Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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